A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a = a₁ + Fm where

Question

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a = a₁ + Fm where a₁ = 3.00 meter/second², F = 12.0 kilogram⋅meter/second² and m =7.00 kilogram. First, which of the following is the correct step for obtaining a common denominator for the two fractions in the expression in solving for a?
a. (m/m times a₁/1) + (1/1 times F/m)
b. (1/m times a₁/1) + (1/m times F/m)
c. (m/m times a₁/1) + (F/F times F/m)
d. (m/m times a₁/1) +(m/m times F/m )

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Ngọc Diệp 1 week 2021-07-22T09:49:45+00:00 1 Answers 1 views 0

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    2021-07-22T09:51:37+00:00

    Answer:

    The correct option is option (a).

    The acceleration of an object is \frac{33}{7} m/s².

    Explanation:

    Given expression is

    a=a_1+\frac Fm

    [ divide a whole number by 1 to turn into a fraction. Since a_1 is a whole number, so  a_1  is divided by 1 to turn into a fraction]

    \Rightarrow a=\frac{a_1}{1}+\frac Fm

    [The l.c.m of the denominators 1 and m is m. Now multiply both numerator and denominator by m of \frac{a_1}1 and multiply both numerator and denominator by 1 of \frac Fm]

    \Rightarrow a=(\frac mm\times\frac{a_1}{1})+(\frac 11\times\frac Fm)

    The correct option is option (a)

    Given that,

    F= 12.0 kg.m/s² , m=7.00 kg and a_1 = 3.00 m/s²

    \therefore a=(\frac mm\times\frac{a_1}{1})+(\frac 11\times\frac Fm)

         =(\frac{7.00}{7.00}\times \frac{3.00}{1})+(\frac11\times \frac{12.0}{7.00})

        =\frac{21}{7}+\frac{12}{7}

        =\frac{21+12}{7}

        =\frac{33}{7} m/s²

    The acceleration of an object is \frac{33}{7} m/s².

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